For each problem provide the following: . Hypotheses Pre-test results and transf
ID: 3360771 • Letter: F
Question
For each problem provide the following: . Hypotheses Pre-test results and transformations (if necessary) Appropriate ANOVA table Post-test (if applicable) Conclusion: statistical & biological . . You are measuring weights (Kg) of sea lion pups from five areas, Santa Barbara Isl. Santa Rosa Isl, the Farallons, Monterey and a beach in Baja Mexico. 5 pups from each location were measured, but on the way home, a sea gull stole the data sheet for one of the pups from Monterey. Your advisor won't let you go back and remeasure another sea lion. From the data below, are the weights from these populations different? Santa Barbara Santa Rosa Farallons 27.61 29.60 25.81 19.91 24.70 16.30 16.37 23.17 28.32 24.79 26.24 22.38 16.21 14.241345 14.45 16.00 16.94 14.92 15.96 14.82 16.79 15.56Explanation / Answer
Hypothesis of the data,
we are given above the table of ANOVA,
The hypotheses of interest in an ANOVA are as follows:
where k = the number of independent comparison groups.
the formulas for calculation for ANOVA,
df
ss
mss
f
p
Between
k-1
SSb
MSB=SSB/k-1
F=MSB/MSE
within
n-k
SSE
MSE=SSE/n-k
total
n-1
SST
the analysis table for anova is
Source of Variation
SS
Df
MS
F
P-value
F crit
Between Groups
344.428
4
86.10701
5.545853
0.004334
2.927744
Within Groups
279.4748
18
15.52638
Total
623.9028
22
conclusion:
p-value is =0.004, p<0.5 therefor we reject the null hypothesis sis and we can conclude that the there is significant difference between the all groups.
Post test (Tukeys test)
The p-value corresponding to the F-statistic of one-way ANOVA is lower than 0.05 which strongly suggests that
one or more pairs of treatments are significantly different. You have k=4k=4 treatments, for which we shall apply
Tukey's HSD test to each of the 6 pairs to pinpoint which of them exhibits statistically significant difference.
We first establish the critical value of the Tukey-Kramer HSD QQ statistic based on the k=5 treatments
and =18 degrees of freedom for the error term, for significance level =0.05 (p-values) in
the Student zed Range distribution. We obtain these critical values for Q for of 0.05,
critical value of Q at Q =0.05,k=5,=18 Q= 4.27
treatments
pair
Tukey HSD
Q statistic
Tukey HSD
p-value
Tukey HSD
inferfence
A vs B
2.1201
0.5684622
insignificant
A vs C
1.5753
0.7750580
insignificant
A vs D
3.4629
0.1473794
insignificant
A vs E
3.4161
0.1562257
insignificant
B vs C
3.6954
0.1098259
insignificant
B vs D
5.4618
0.0088349
** p<0.01
B vs E
5.4150
0.0094772
** p<0.01
C vs D
1.9777
0.6224585
insignificant
C vs E
1.9309
0.6402139
insignificant
D vs E
0.0444
0.8999947
insignificant
So
Biological conclusion,
thanks.
df
ss
mss
f
p
Between
k-1
SSb
MSB=SSB/k-1
F=MSB/MSE
within
n-k
SSE
MSE=SSE/n-k
total
n-1
SST
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